Solution - Career Point

For destuctive interferenceI min = ( I ) 21 − I 2 = ( 2I− I ) 2ThenIIminmax=((2I −2I +25. [B] Using decay equationA 2 = A 1 e –λtor e –λt =time t =(4πa)26. [B] V x = ka= 4π⎛⎜Q k =⎝I)I)A 2Aor λt = log1A1A 222⎛ 2 1⎞= ⎜−⎟1= 1 2⎝ + ⎠ 34log A 1 / A 2⎛ A ⎞= 1.44T log eλ⎜1⎟ ⎝ A 2 ⎠2 σ14πε2 σk(4πb)–b14πε σ= (a – b + c)ε 027. [D] Let the given quantity be x 1 then,0⎞⎟⎠x = 3π (a 2 – b 2 )h = 3π (a 2 h – b 2 h)= 3π a 2 h – 3π b 2 h022 σk(4πc)+bσ(a – b + c)Each term has the dimension of x 1 then[x] = [a 2 h] = [L 2 L] = [L 3 ] and also[x] = [b 2 h] = [L 2 L] = [L 3 ] quantity is volume.A x U(x2 + B)28. [B] U = or A =x2 + BxHere dimensions of x 2 and B should be same.i.e. [B] = [x 2 ] = [L 2 ]⎡2 −2ML T ⎤Also [A] = ⎢1/ 2⎥ [L 2 ] = [ML 7/2 T –2 ]⎢⎣L ⎥⎦Then [AB] = [ML 7/2 T –2 ] [L 2 ] = [ML 11/2 T –2⎛ t ⎞ dx ⎞29. [A] v = u ⎜1 − ⎟ or = u ⎜⎛ t1 − ⎟⎠⎝ t´⎠ dt ⎝ t ´⎛2⎞integrating, x = u ⎜tt − ⎟ + C⎝ 2t´⎠at t = 0, n = 0 and c = 0⎛2⎞∴ x = u ⎜t⎟⎛ t ⎞t − = 10t ⎜ t − ⎟⎝ 2t´⎠ ⎝ 10´⎠Putting t = 10⎛ 10 ⎞x = 10 × 10 ⎜1 − ⎟ = 0⎝ 10 ⎠30. [A] using h = 21 gt 2 , we get t 1 =2hglet t 1 be the time taken from instants of jumpingto the opening of parachute, then2× 40t 1 = = 2.86 sec9.8His velocity at this point is given byv 2 1 = 2gh 1 = 2 × 9.8 × 40= 784 or v 1 = 28 ms –1for the remaining journey,v = v 1 + at 2v − u 2 − 28or t 2 = = = 13seca − 2∴ total time = t 1 + t 2 = 2.86 + 13= 15.86 ≅ 16 s31. [B] Let u be the velocity of projectile w.r.t. tanksvelocity v thenu x = u cos 30 + v; u y = u sin 30º2u sin 30ºand T =g2u sin 30ºRange, R 1 = u x T = (u cos 30º + v)gfor y axisu x´ = u cos 30º – v and u y´ = u sin 30º2u sin 30ºT =gRange e, R 2 = Tu´x2u sin 30º= (u cos 30º – v)g4u2Then R 1 + R 2 = (sin 30º cos 30º)g4uR 1 – R 2 = v sin 30ºgEliminating u we getv 2 ==g4 tan 30104 tan 30⇒ 4.9 ms –1(R1− R 2 )(R + R )1222(250 − 200)= 24 m 2 s –2(250 + 200)32. [D] Let α be the angle between velocities of pair ofparticles then relative velocity is given by2 2αv r = v + v − 2v×v×cosα= 2 v sin 22π2v(sin α / 2) 4average v r =∫dα =2πv π0dα∫0XtraEdge for IIT-JEE 112APRIL 2010